Middle Term, Greatest Term, Greatest Coefficient: Binomial Theorem Made Clear
You expand (1 + x)¹⁰ in your head and someone asks, "Which term is the biggest?" Most students start writing out all eleven terms. You do not need to: the terms rise, peak and fall, and a single ratio tells you exactly where the peak sits. This post covers three related questions that JEE Main and Advanced keep asking: the middle term, the greatest term, and the greatest coefficient. These are Class 11 Binomial Theorem ideas and matter for JEE only.
Quick Summary
(a + b)ⁿ has n + 1 terms. If n is even there is one middle term, T(n/2 + 1). If n is odd there are two, T((n+1)/2) and T((n+3)/2).
The general term is T(r+1) = ⁿCr · aⁿ⁻ʳ · bʳ, and the ratio T(r+1) / T(r) = [(n − r + 1) / r] × (b / a).
Greatest coefficient depends only on n. Greatest term depends on the actual value of x as well.
Greatest term: compute m = (n + 1)|b/a| / (1 + |b/a|). If m is not an integer, the greatest term is T(⌊m⌋ + 1). If m is an integer, T(m) and T(m + 1) are equal and both greatest.
Start from what you know
You already know the general term of (a + b)ⁿ: T(r+1) = ⁿCr · aⁿ⁻ʳ · bʳ, where ⁿCr = n! / (r!(n − r)!) is the number of ways to choose r items from n, and r runs from 0 to n. Counting from r = 0 is why the term number is always one more than the power of b.
The middle term
There are n + 1 terms in all. If n is even, n + 1 is odd, so exactly one term sits in the centre: the one with r = n/2, that is T(n/2 + 1). If n is odd, n + 1 is even, so two terms share the centre: r = (n−1)/2 and r = (n+1)/2, which are T((n+1)/2) and T((n+3)/2).
Example: the middle term of (x + 1/x)¹⁰. Here n = 10 is even, so the middle term is T₆ with r = 5. T₆ = ¹⁰C₅ · x⁵ · (1/x)⁵ = 252 · x⁵ · x⁻⁵ = 252. The powers of x cancel completely, so the middle term here is the constant term. This is a classic exam link: for (x + 1/x)ⁿ the middle term is constant, and for n = 10 it is 252.
The greatest coefficient
Look at the row of ⁿCr values. They are symmetric (ⁿCr = ⁿCn−r) and they rise to the centre. So the largest binomial coefficient is the middle one: ⁿCn/2 when n is even, and ⁿC(n−1)/2 = ⁿC(n+1)/2 when n is odd. For n = 9 the two largest are both 126; for n = 10 the single largest is 252.
The greatest term: why the ratio works
A term is bigger than the one before it exactly when their ratio exceeds 1. Divide consecutive terms:
T(r+1) / T(r) = [ⁿCr / ⁿCr−1] × (b/a) = [(n − r + 1) / r] × (b / a).
As r increases, the fraction (n − r + 1)/r falls steadily, so the terms grow at first and then shrink. The peak is where the ratio crosses 1. Set the ratio ≥ 1 with k = |b/a| (we compare sizes, so we ignore signs):
(n − r + 1) k ≥ r, so r ≤ (n + 1)k / (1 + k).
Call the right side m. The term T(r+1) is at least as large as T(r) while r ≤ m. So the last increase happens at r = ⌊m⌋, making T(⌊m⌋ + 1) the greatest. If m is itself a whole number, then at r = m the ratio is exactly 1, so T(m) and T(m + 1) are equal and both are greatest.
Worked example
Find the greatest term in the expansion of (1 + 2)¹⁰, treated as (1 + x)¹⁰ with x = 2.
Here n = 10 and k = |b/a| = 2.
m = (n + 1)k / (1 + k) = 11 × 2 / 3 = 22/3 ≈ 7.33.
m is not an integer, so the greatest term is T(⌊7.33⌋ + 1) = T₈, with r = 7.
T₈ = ¹⁰C₇ × 2⁷ = 120 × 128 = 15360.
Check the neighbours. T₇ = ¹⁰C₆ × 2⁶ = 210 × 64 = 13440. T₉ = ¹⁰C₈ × 2⁸ = 45 × 256 = 11520. Both are smaller than 15360, so T₈ is indeed the greatest.
Now the integer case: (1 + x)⁹ with x = 1. Here m = 10 × 1 / 2 = 5, a whole number. So T₅ and T₆ tie: ⁹C₄ = ⁹C₅ = 126. That matches the "two greatest coefficients when n is odd" result, because with x = 1 terms equal coefficients.
Mistakes students actually make
Confusing coefficient with term. The greatest coefficient of (1 + 2x)¹⁰ is not the greatest term. The binomial coefficient is largest at the middle, but the powers of 2 push the greatest term to the right.
Taking the term number as r. The formula gives r; the term is T(r + 1). Forgetting the +1 gives an off-by-one answer every time.
Dropping the factor b/a. For (3 + 2x)ⁿ the ratio uses k = |2x/3|, not |x|. Always rewrite the ratio with the real a and b.
Applying the middle-term rule to the greatest term. They coincide only when |b/a| = 1.
Missing the tie. When m is an integer, two terms are greatest; answering with one loses the mark.
How it is asked in JEE
"Find the middle term(s) of (a + b)ⁿ" with a given n, often with fractional or negative bases.
"If the greatest term in the expansion of (1 + x)ⁿ for a given x is Tk, find n or x", where you must work backwards from the inequality for m.
Numerical-value questions asking for the value of the greatest coefficient or of the term itself.
Combined questions where the middle term equals a constant, as with (x + 1/x)ⁿ.
Pro Tip: Never memorise "the greatest term is the middle term". Memorise the ratio T(r+1) / T(r) = [(n − r + 1) / r] × (b / a) and set it ≥ 1. It derives the m formula in one line, works for any a and b, and handles ties automatically. When we teach this at Sukriti NeetJee, the first thing students are asked to do is derive the ratio, because it is the one idea that rebuilds every other result.
Frequently Asked Questions
How many middle terms does (a + b)ⁿ have?
One if n is even, namely T(n/2 + 1), and two if n is odd, namely T((n+1)/2) and T((n+3)/2). This comes from the fact that there are n + 1 terms in total.
Is the greatest term always the middle term?
No. It is the middle term only when |b/a| = 1. For other values the peak shifts right if |b/a| > 1 and left if |b/a| < 1.
What is the greatest coefficient in the expansion of (1 + x)ⁿ?
It is ⁿCn/2 when n is even, and ⁿC(n−1)/2 (equal to ⁿC(n+1)/2) when n is odd. It depends only on n, not on x.
What do I do if the base has a negative sign, like (1 − 2x)ⁿ?
Compare magnitudes only. Use k = |b/a| in the formula for m to locate the greatest term in absolute value. The sign of that term then follows from (−1)ʳ.
What if m comes out as a whole number?
Then T(m) and T(m + 1) are equal and both are the greatest term. The ratio equals exactly 1 at r = m, so there is no strict winner.
Stuck on a greatest-term problem? Drop your doubt in the comments, and share this with a fellow JEE aspirant who still expands every term by hand.
Comments